# Electronic home thermostats. Whats the best temp to set it at?

Sailor - So is the example I gave out earlier accurate? That no fuel savings would occur if the temperature was always either falling or rising and a ‘lower limit’ holding temperature was never reached?

I ask because this is the situation I have currently in my house. I set back the temp from 70 to 62 while at work. However over the holidays I observed the temperature would only fall to roughly 64 degrees before the evening heating kicked in (now I assume that when winter finally arrives in Chicago - like today!- the rate of decline will increase and the house will probably hit 62 degrees during the daytime).

No one’s questioning your background or the veracity of your information. But usually “trust me” is not a thread-ender on this board. People are interested in more than a bottom-line answer–the discussion and underlying information is why people use this board. I’m terribly sorry if I’ve missed other similar discussions.

No, you can forget about the furnace. The heat loss is directly proportional to three things:

1- Thermal conductivity (or the reverse: insulation) which is constant
2- Temperature difference between inside and outside
3- Time.

Whether the temperature is falling or rising is irrelevant, what counts is the temperature inside: the higher the temperature the higher the amount of heat lost which will have to be replaced. So, you do save by letting the temperature drop, even if it dos not drop much.

Obviously something is wrong here, because if you had instead set the house at 66 degrees, your interpretation says that you would get a fuel savings. You are saving energy whenever the house is below 70 degrees, because less heat will leave the house, since the amount of heat leaving is proportional to the temperature difference.

As to how obvious it is that lower temperature = energy savings, it’s not of necessity the case that lower is always better. If the furnace somehow became less efficient when run non-stop for a long time, you’d have to take that into account. (I suspect the furnace actually becomes more efficient, based on pipper’s reasoning.)

In the similar case of setting the temperature higher when air conditioning during the summer, I’ve heard setting too low can be bad, because when an air conditioner runs non-stop for a long time, it can ice up, reducing the efficiency. Whether or not that actually occurs for any given air conditioner, it’s certainly plausible, and can’t be discounted as easily.

ZenBeam, I could find a plausible scenario where driving your car at 80MPh would save gas compared to driving at 50 MPh. I suppose I could find a plausible scenario where burning \$100 bills to heat yourself would make sense too. But that’s not realistic advice in general.

Couldn’t this be settled by a simple experiment. Find a couple of days where the weather is forcast to be the same – same cloud cover, daytime temp, etc. On day one, leave the thermo at a comfortable temp all day and record how much electricity (and/or gas)went through the meter. On the second day, lower the daytime thermo and see if a different amount of fuel passed through the meter.

In warmer regions, does the same rule about heat-loss (or gain) apply when air-conditioning the house? Is it better to warm the house during the day, then bring it back to cool after work?

Well, I’m an engineer, and I can do the first part (I think) - determining how much energy is lost by a house maintained at a constant temperature. What I can’t do right away is to determine the loss from a house where the house temperature falls naturally. I would have to make some more assumptions about the heat capacity of the entire house. Maybe I can come back later, or someone else can take it the next step.

Assume: A 30x40 house with 12-foot exterior walls, flat roof.

Total area: 2*(30x12) + 2*(4012) + 3040 = 2880 square feet

Assume: No slab loss.

Assume: R-19 insulation all-around. This is 19 ft[sup]2[/sup]*F/Btu/hr thermal resistance.

Assume: Measuring from 7:00 am to 5:00 pm - 10 hours total.

Assume: Exterior temperature starts at 20 F, rises to 40 F by 3:00, falls to 30 F by 5:00. Smooth linear rise and fall.

Map: Exterior temperatures

7:00am - 20F
8:00am - 22.5F
9:00am - 25F
10:00am - 27.5F
11:00am - 30F
12:00pm - 32.5F
1:00pm - 35F
2:00pm - 37.5F
3:00pm - 40F
4:00pm - 35F
5:00pm - 30F

Equation to determine heat loss: Q/R

Q = Area*(Tinside - Toutside)
R = 19 ft[sup]2[/sup]*F/Btu/hr

Loss is given in terms of Btu/hr.

Assumption 1: House interior maintained at constant 70F

7-8am: Energy = 2880 ft[sup]2[/sup](70-21.25)/19 = 7389.5 Btu
8-9am: Energy = 2880
(70-23.75)/19 = 7010 Btu
9-10am: Energy = 2880*(70-26.25)/19 = 6332 Btu
10-11am: Energy = 2880*(70-28.75)/19 = 6253 Btu
11-12pm: Energy = 2880*(70-31.25)/19 = 5874 Btu
12-1pm: Energy = 2880*(70-33.75)/19 = 5495 Btu
1-2pm: Energy = 2880*(70-36.25)/19 = 5116 Btu
2-3pm: Energy = 2880*(70-38.75)/19 = 4737 Btu
3-4pm: Energy = 2880*(70-37.5)/19 = 4926 Btu
4-5pm: Energy = 2880*(70-32.5)/19 = 5684 Btu

Total Energy Required: 58,816.5 Btu

Look OK? Now…I would do the example where the house temperature is allowed to fall, if we can determine a specific heat capacity of the house and its belongings, to determine how fast the interior temperature drops.

You can experiment and the results will confirm what I said or the experiment was done wrong. We are talking about something pretty basic here. There are plenty of people out there doing experiments and getting results which contradict the most basic laws of physics. That does not mean their experiments are correct.

Yes, the same rule applies. The warmer the house at any given moment, the more you save. There is no "it costs more to cool it later. It also applies to water heaters and any other similar instance.

Optimal settings aside, I will say that I just purchased a new Honeywell programmable thermostat that I think is excellent. It has a modifiable feedback loop that gives it a type of “learning ability”. Over the course of a few days it recognizes how quickly your house increases/decreases temperature and adjusts itself to turn the furnace on/off at the proper time. So you no longer have guess that turning the thermostat back up at 5:30 should result in it being warm by 6:00- the unit learns this on its own. It constantly monitors the rate of temperature change which allows it to gradually accomodate for decreasing/increasing outdoor temperatures. Probably not as big a deal for forced air systems that make a house feel warmer as soon as they kick in, but it certainly has been a lot nicer for our slower radiator system (it also seems to regulate a constant temperature better and doesn’t overshoot the setting like the old themostat).

What if you used a heat pump instead of gas? Would it make sense to keep the house warmer in the daytime (when it is cheaper to pump heat into the house because the ambient temperature is relatively high) and count on retention of that daytime cheap heat to keep you cozy all night or would it be better to use the same strategy as you would for a gas furnace where you would do most of your heating at night.

(Actually, the real question I want the answer to is what strategy to use in the summer with air conditioning.)

Actually you don’t want to use a set-back strategy with a heat pump. Most heat pumps have an inefficient secondary heating element (usually electrical) that helps boost the temps quickly. If you set the themostat down, when it comes time to reheat the house you’ll be doing it with the secondary element which ends up being more costly than having let the heat pump maintain a constant temp.

Anthracite, I believe you are making it more complex than it needs to be. A house loses heat directly in proportion to the temperature difference with the outside and the conductivity. It is a simple differential equation similar to the discharge of a capacitor across a resistor. (I am not going to attempt posting formulas with sub and superindices here).

The furnace would be represented by a current source so, the mathematical model is extremely simple and there’s no need to specify specific values of temperatures or insulation. It is easy to design an electronic circuit which has the same behavior and then you can just observe it.

An extra complication is when you have hydronic heating as this inserts an extra delay in the system which really complicates it. That was my case. When you set the thermostat up the furnace starts, The water in the radiators gets very hot before the room temperature gets up to the set point. Then, even though the furnace shuts off, the water continues to transmit heat out. The mathematical model is a bit more complex but not much. These are simple problems of feedback and control systems.

OK, I made some simplifying assumptions. I hope they are not wrong.

Assumption 2: House interior allowed to cool to 50 F until 4:00pm. Then is increased to 70F over one hour. Assume (nearly) constant drop of interior temps over the day.

OK…let’s assume that 1/10 of the volume of the house is wood, and 9/10 air.

Total house volume is therefore 304012 = 14,400 ft[sup]3[/sup]

Wood, at 36 lbm/ft[sup]3[/sup], makes up 1/30 of that, or 480 ft[sup]3[/sup]. This equates to 17,280 lbm wood.

Air, at 0.063 lbm/ft[sup]3[/sup], makes up 29/30 of that, or 13,920 ft[sup]3[/sup]. This equates to 878 lbm air.

At specific heat capacities of 0.4 Btu/lbmR for wood and 0.24 Btu/lbmR for air, we have a house heat capacity of:

17,280 lbm * 0.4 Btu/lbmR + 878 lbm * 0.24 Btu/lbmR = 7123 Btu/R.

Map: Exterior temperatures / Interior Temperatures

Since I’m not going to do the differential equation, I’ll guesstimate. I’ll assume that the temperature drop of the house is assessed at the end of each hour.

7-8am: Energy = 2880 ft[sup]2[/sup]*(70-21.25)/19 = 7389.5 Btu

New house temperature: 70 - (7389.5 / 7123) = 68.96 F

8-9am: Energy = 2880*(68.96-23.75)/19 = 6853 Btu

New house temperature: 68.96 - (6853 / 7123) = 68 F

9-10am: Energy = 2880*(68-26.25)/19 = 6328 Btu

New house temperature: 68 - (6328 / 7123) = 67.1 F

10-11am: Energy = 2880*(67.1-28.75)/19 = 5813 Btu

New house temperature: 67.1 - (5813 / 7123) = 66.3 F

11-12pm: Energy = 2880*(66.3-31.25)/19 = 5310 Btu

New house temperature: 66.3 - (5310 / 7123) = 65.6 F

12-1pm: Energy = 2880*(65.6-33.75)/19 = 4821 Btu

New house temperature: 65.6 - (4821 / 7123) = 64.9 F

1-2pm: Energy = 2880*(64.9-36.25)/19 = 4346 Btu

New house temperature: 64.9 - (4346 / 7123) = 64.3 F

2-3pm: Energy = 2880*(64.3-38.75)/19 = 3871 Btu

New house temperature: 64.3 - (3871 / 7123) = 63.75 F

3-4pm: Energy = 2880*(63.75-37.5)/19 = 3979 Btu

New house temperature: 63.75 - (3979 / 7123) = 63.2 F

Now: It’s 4:00pm. We need to heat the house to 70 F. This means we need to regain the heat we lost, and account for continued heat losses over the hour. First - the heat losses over the hour of heating:

4-5pm: Energy = 2880*(((70+63.2)/2) - 32.5)/19 = 5168 Btu

Now, add to that the heat required to raise the house from 63.2 F to 70 F:

(70-63.2)*7123 = 48,436 Btu

48,436 + 5168 = 53,604 Btu

Which is less heat required than the previous example.

Please check for math and logic errors. I’m in a hurry, and may have screwed up.

I would like to see some confirmation of this. AFAIK the the purpose of the resistor heating element is not to gain any heating speed (which the system does not know if you want or need anyway) but to heat when the temperature outside is so low that the pump cannot function efficiently. In anycase, if enough time elapses, it would still be worth while. If you give me the specs of any heat pump system which does this, I can do the calculations, but I have never seen this.

I frankly don’t know all that much about heat pumps and I’ve just heard this as a rule of thumb. I have never had the opportunity to deal with a heat pump to heat a building.

For those too lazy to search:

“Where should I set the day thermostat temp for max efficiency?”